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Classical Model
In this model, the stability of a single generator connected to aninfinite bus is analyzed.
Though this model is not suitable for current power systems, it helpsus understand the basic phenomenon of stability.
Assumptions
1 Exciter dynamics are neglected and the filed current is assumed to beconstant so that the induced voltage is always constant.
2 Damper winding dynamics are neglected.
3 The mechanical input power is assumed to be constant during theperiod of study.
4 Rotor is assumed to be of cylindrical type so that no saliency ispresent.
Siva (IIT P) EE549 1 / 61
SMIB System
Let us consider Single Machine Infinite Bus (SMIB) system.
E δ V∞ = 1∠0◦
Pe =EV sin δ
X
Where X = Xg + XTr + XTL in p.u.
Siva (IIT P) EE549 2 / 61
Rotor Dynamics - Swing EquationThe equation governing rotor motion of a synchronous machine is given as
Jd2θmdt2
= Ta = Tm − Te N-m
whereJ = the total moment of inertia of the rotor masses in kg-m2
θm = the angular displacement of the rotor with respect to a stationaryaxis in mechanical radians (rad)t = time in seconds (s)Tm = the mechanical or shaft torque supplied by the prime mover in N-mTe = the net electrical or electromagnetic torque in N-mTa = the net accelerating torque in N-m
Siva (IIT P) EE549 3 / 61
Tm and Te are considered positive for the synchronous generator.
Tm accelerates the rotor in the positive θm in the direction of rotation.
For a motor, Tm and Te are reversed in sign.
In the steady state, Tm = Te . Hence, Ta = 0.
θm is measured with respect to a stationary reference axis on the stator. Torepresent it with respect to the synchronously rotating frame, let us define
θm = ωsmt + δm
whereωsm is the synchronous speed of the machine in mechanical radians persecondδm is the angular displacement of the rotor in mechanical radians from thesynchronously rotating reference axis.
Siva (IIT P) EE549 4 / 61
dθmdt
= ωsm +dδmdt
ωm − ωsm =dδmdt
where ωm =dθmdt
is the angular velocity of the rotor in mechanical radians
per second.Differentiating it again,
d2θmdt2
=d2δmdt2
Substituting it,
Jd2δmdt2
= Ta = Tm − Te N-m
On multiplying by ωm,
Jωmd2δmdt2
= Pa = Pm − Pe
Siva (IIT P) EE549 5 / 61
wherePm = shaft power input in MWPe = electrical power output in MWPa = accelerating power in MW
Let us define inertia constant H.
H =stored kinetic energy in megajoules at synchronous speed
Machine rating in MVA
H =12Jω
2sm
SmachMJ/MVA
Substituting it,2H
ω2sm
ωmd2δmdt2
=Pa
Smach=
Pm − Pe
Smach
In practice, ωm does not differ significantly from the synchronous speed.ωm ≈ ωsm
2H
ωsm
d2δmdt2
= Pa = Pm − Pe per unit
Siva (IIT P) EE549 6 / 61
It can be written as
2H
ωs
d2δ
dt2= Pm − Pe per unit
δ and ωs have consistent units which can be mechanical or electricaldegrees or radians.
This equation is called the swing equation of the machine.
It is a second-order nonlinear differential equation.
When it is solved, we obtain δ as a function of t. This is called theswing curve.
Siva (IIT P) EE549 7 / 61
It can be written as two first-order differential equations.
2H
ωs
dω
dt= Pm − Pe per unit
dδ
dt= ω − ωs
ω, ωs and δ involve electrical radians or electrical degrees.
δ is the load angle.
Siva (IIT P) EE549 8 / 61
Small Signal Stability
δ
P
Pm
Pmax
δ0 δmax
Consider small incremental changes in δ around δ0. The swing equationcan be expressed as follows:
2H
ωs
d2(δ0 + ∆δ)
dt2= Pm − Pmax sin(δ0 + ∆δ)
2H
ωs
d2δ0dt2
+2H
ωs
d2∆δ
dt2= Pm − Pmax sin δ0 cos ∆δ − Pmax cos δ0 sin ∆δ
Siva (IIT P) EE549 9 / 61
Since ∆δ is very small,
sin ∆δ ≈ ∆δ; cos ∆δ ≈ 1
2H
ωs
d2δ0dt2
+2H
ωs
d2∆δ
dt2= Pm − Pmax sin δ0 − Pmax cos δ0∆δ
At the stable equilibrium point (δ0),
2H
ωs
d2δ0dt2
= Pm − Pmax sin δ0 = 0
Therefore,2H
ωs
d2∆δ
dt2= −Pmax cos δ0∆δ
Let Ps = Pmax cos δ0. Ps is called the synchronizing power or torque. Inper unit system, torque and power are equal.
2H
ωs
d2∆δ
dt2+ Ps∆δ = 0
Siva (IIT P) EE549 10 / 61
It is a linear second-order differential equation.
When Ps is positive, the solution ∆δ(t) is an undamped sinusoid.
When Ps is negative, the solution ∆δ(t) increases exponentiallywithout limit.
Ps is positive at δ0 and negative at δmax .
Therefore δ0 is a stable equilibrium point and δmax is an unstableequilibrium point.
Siva (IIT P) EE549 11 / 61
Let us apply the Laplace transformation to it.
2H
ωss2∆δ(s) + Ps∆δ(s) = 0
On solving, we get two roots.
s = ±√−Psωs
2H
If Ps is positive,
s = ±√
Psωs
2H
The system has sustained oscillation. It is marginally stable.If Ps is negative,
s = +
√Psωs
2H,−√
Psωs
2H
The system is unstable.
Siva (IIT P) EE549 12 / 61
In synchronous machines, damper or amortisseur windings are there.
1 They damp out oscillations in generators.
2 They help synchronous motors start as induction motors becausesynchronous motors do not have starting torque.
The damping torque depends on the rate change of rotor angle.
Pd = Ddδ
dt
where D is the damping coefficient. After including this, the linearizedswing equation can be written as
2H
ωs
d2∆δ
dt2+ D
d∆δ
dt+ Ps∆δ = 0
Siva (IIT P) EE549 13 / 61
Applying the Laplace transformation,
2H
ωss2∆δ(s) + Ds∆δ(s) + Ps∆δ(s) = 0
s2 +ωs
2HDs +
ωs
2HPs = 0
By comparing this with the standard characteristics equation of a secondorder system,
s2 + 2ζωns + ω2n = 0
we get
ωn =
√ωs
2HPs
ζ =D
2
√ωs
2HPs
where ωn is the natural frequency of oscillations and ζ is the dampingratio.
Siva (IIT P) EE549 14 / 61
The roots (eigen values) of the characteristics equation are
s = −ζωn ± ωn
√1− ζ2
If Ps is positive
there will be two complex conjugate roots with negative real part.
hence the system will be stable.
If Ps is negative,
there will be two real roots with one positive.
hence the system will be unstable.
Siva (IIT P) EE549 15 / 61
The linearized swing equations can also be written as follows:
2H
ωs
d∆ω
dt+ D∆ω + Ps∆δ = 0
d∆δ
dt= ∆ω
In state space form,[∆ω̇
∆δ̇
]=
[−ωsD
2H−ωsPs
2H1 0
] [∆ω∆δ
]By taking the Laplace transformation,[
s∆ω(s)−∆ω(0)s∆δ(s)−∆δ(0)
]=
[−ωsD
2H−ωsPs
2H1 0
] [∆ω(s)∆δ(s)
]
Siva (IIT P) EE549 16 / 61
[∆ω(s)∆δ(s)
]=
[s +
ωsD
2H
ωsPs
2H−1 s
]−1 [∆ω(0)∆δ(0)
]This can be written as[
∆ω(s)∆δ(s)
]=
[s + 2ζωn ω2
n
−1 s
]−1 [∆ω(0)∆δ(0)
]If the rotor angle is perturbed by a small angle, ∆δ(0) = ∆δ and∆ω(0) = 0.
∆ω(s) = − ω2n∆δ
s2 + 2ζωns + ω2n
∆δ(s) =(s + 2ζωn)∆δ
s2 + 2ζωns + ω2n
Siva (IIT P) EE549 17 / 61
By taking the inverse Laplace transformation, we can get the following.
∆ω(t) = − ωn∆δ√1− ζ2
e−ζωnt sinωd t
and
∆δ(t) =∆δ√1− ζ2
e−ζωnt sin(ωd t + θ)
where ωd = ωn
√1− ζ2 and θ = cos−1(ζ).
The motion of rotor relative to the operating point is
δ = δ0 +∆δ√1− ζ2
e−ζωnt sin(ωd t + θ)
and the rotor angular frequency is
ω = ω0 −ωn∆δ√1− ζ2
e−ζωnt sinωd t
Siva (IIT P) EE549 18 / 61
The time constant is
τ =1
ζωn
τ =4H
ωsD
The response approximately settles at 4 to 5 time constants.
As H increase, it results in a longer settling time.
As Ps increases, it results in an increase in ωn.
Siva (IIT P) EE549 19 / 61
Example
A 50 Hz synchronous generator having inertia constant H = 5 sec and adirect axis reactance X ′d = 0.3 p.u. is connected to an infinite bus througha transformer and a double circuit line. The network is purely reactive. Thesynchronous generator is delivering real power P = 0.8 p.u. and reactivepower Q = 0.074 p.u. to the infinite bus of 1.0 p.u at steady state.
0.2
E ′d
X ′d = 0.3
V∞ = 1∠0◦
X = 0.3
X = 0.3
Assume the per unit damping power coefficient D = 0.2. Consider a smalldisturbance of ∆δ = 10◦. For example, the breakers open and quickly close.Determine the motion of rotor angle and the generator frequency.
Siva (IIT P) EE549 20 / 61
The current flowing into the infinite bus is
I =S∗
V ∗=
0.8− 0.074
1∠0◦= 0.8− 0.074
The reactance between E ′d and V∞ before the fault is
X = 0.3 + 0.2 +0.3 + 0.3
2= 0.65
The direct axis transient internal voltage is
E ′d = V∞ + X1I = 1∠0◦ + 0.65× (0.8− 0.074) = 1.17∠26.38◦
Pmax =E ′dV∞X1
=1.17× 1
0.65= 1.8
δ0 = 26.38◦
Siva (IIT P) EE549 21 / 61
Ps = Pmax cos δ0 = 1.8× cos(26.38◦) = 1.6125
ωn =
√ωs
2HPs = 7.1174 rad/s
fn = 1.133 Hz
ζ =D
2
√ωs
2HPs= 0.4414
ωd = ωn
√1− ζ2 = 6.3866 rad/s
θ = cos−1 ζ = 92.4◦
δ = 26.38◦ + 11.1444e−3.1416t sin(6.4t + 92.4◦)
f = 50− 0.2525e−3.1416t sin 6.4t
Siva (IIT P) EE549 22 / 61
t (s)
δ (deg)
26.38◦
Figure: Swing Curve
t (s)
f (Hz)
50
Figure: Frequency
Siva (IIT P) EE549 23 / 61
Transient Stability
The stability of a system is analyzed when it is subjected to largedisturbances like faults.
Since the disturbances are large, the rotor angle will swing high.
This does not let us linearize the swing equation.
The analysis of large disturbance stability is difficult.
The equation has to be solved using numerical integration techniques.
However, for an SMIB system, a direct approach called Equal AreaCriterion method can be used to analyze it.
It is a graphical approach.
Siva (IIT P) EE549 24 / 61
Equal Area Criterion
2H
ωs
d2δ
dt2= Pm − Pe
d2δ
dt2=ωs
2H(Pm − Pe)
Let us multiply both sides of the above equation by 2dδ/dt,
2dδ
dt
d2δ
dt2=ωs(Pm − Pe)
H
dδ
dt
d
dt
[dδ
dt
]2=ωs(Pm − Pe)
H
dδ
dt
On integration, [dδ
dt
]2=
∫ δm
δ0
ωs(Pm − Pe)
Hdδ
Siva (IIT P) EE549 25 / 61
For a system to be stable,dδ
dt= 0 after a disturbance.
∫ δm
δ0
ωs(Pm − Pe)
Hdδ = 0
∫ δm
δ0
(Pm − Pe)dδ = 0
where
δ0 = initial rotor angle
δm = maximum rotor angle
Siva (IIT P) EE549 26 / 61
Sudden change in Pm
δ
P
Pm0
Pm1
Pmax
δ0δ1δm
Siva (IIT P) EE549 27 / 61
∫ δm
δ0
(Pm1 − Pe)dδ = 0
δ0 is the initial rotor angle. δm is the maximum rotor angle duringoscillation. ∫ δ1
δ0
(Pm1 − Pe)dδ +
∫ δm
δ1
(Pm1 − Pe)dδ = 0
∫ δ1
δ0
(Pm1 − Pe)dδ =
∫ δm
δ1
(Pe − Pm1)dδ
Therefore for the system to be stable
Area(A1) = Area(A2)
Energy Gained = Energy Lost
If A1 > A2, the system will be unstable.
Siva (IIT P) EE549 28 / 61
t
δδm
δ0
δ1
Figure: Swing Curve
Damper or amortisseur windings damp out oscillations.
Siva (IIT P) EE549 29 / 61
Short Circuit Faults
E δ V∞ = 1∠0◦
F
At point F , a three phase fault occurs. To analyze this, we need tounderstand the physical conditions before, during and after the fault.
Siva (IIT P) EE549 30 / 61
1 Before Fault :
−+
E δ
X ′d XTr
−+
1∠0◦
XTL
XTL
X1 = X ′d + XTr +XTL
2
Pe1 = Pmax1 sin δ
where
Pmax1 =EV
X1
Siva (IIT P) EE549 31 / 61
2 During Fault :
−+
E δ
X ′d XTr
−+
1∠0◦
XTL
XTL2
XTL2
The total reactance X between two nodes can be found using Y −∆conversion.X2 during fault will be higher than before fault X1.
Hence,
Pe2 = Pmax2 sin δ
where
Pmax2 =EV
X2
Siva (IIT P) EE549 32 / 61
3 After Fault :
−+
E δ
X ′d XTr
−+
1∠0◦
XTL
X3 = X ′d + XTr + XTL
Pe3 = Pmax3 sin δ
where
Pmax3 =EV
X3
Siva (IIT P) EE549 33 / 61
δ
P
Pm
Pmax1
Pmax3
Pmax2
δ0 δcr δmax
A1
A2
δmax = π − sin−1( PmPmax3
).For the system to be stable, A1 = A2. There is a critical clearing angle δcrbefore which the fault has to be cleared.∫ δcr
δ0
(Pm − Pmax2 sin δ)dδ =
∫ δmax
δcr
(Pmax3 sin δ − Pm)dδ
Siva (IIT P) EE549 34 / 61
Integrating and simplifying the above equation, we get
cos δcr =Pm(δmax − δ0) + Pmax3 cos δmax − Pmax2 cos δ0
(Pmax3 − Pmax2)
t
δ
δm
δcr
tcr
δ0
Figure: Swing Curve
tcr is the critical clearing time in seconds.It will settle at δnew = sin−1( Pm
Pmax3) if damping is present.
Siva (IIT P) EE549 35 / 61
Short Circuit Faults at the end of Transmission Lines(Near the bus)
E δ V∞ = 1∠0◦
F
1 Before Fault :
X1 = X ′d + XTr +XTL
2
Pe1 = Pmax1 sin δ
where
Pmax1 =EV
X1
Siva (IIT P) EE549 36 / 61
2 During Fault :
Since the fault is near the bus, the bus voltage is zero.The power transfer during fault is zero.
Hence,
Pe2 = 0
3 After Fault :
X3 = X ′d + XTr + XTL
Pe3 = Pmax3 sin δ
where
Pmax3 =EV
X3
Siva (IIT P) EE549 37 / 61
δ
P
Pm
Pmax1
Pmax3
δ0 δcr δmax
A1
A2
δmax = π − sin−1( PmPmax3
).For the system to be stable, A1 = A2.∫ δcr
δ0
Pmdδ =
∫ δmax
δcr
(Pmax3 sin δ − Pm)dδ
Siva (IIT P) EE549 38 / 61
Integrating and simplifying the above equation, we get
cos δcr =Pm(δmax − δ0) + Pmax3 cos δmax
Pmax3
We can find the critical clearing time for this case as follows:
d2δ
dt2=ωs
2H(Pm − Pe)
Since Pe = 0 during fault,
d2δ
dt2=ωs
2HPm
Integrating this,dδ
dt=
∫ t
0
ωs
2HPm =
ωs
2HPmt
On further integration,
δ =ωs
4HPmt
2 + δ0
Siva (IIT P) EE549 39 / 61
If δ = δcr ,
δcr =ωs
4HPmt
2cr + δ0
tcr =
√4H(δcr − δ0)
ωsPm
Siva (IIT P) EE549 40 / 61
Factors Influencing Transient Stability
1 How heavily the generator is loaded.
2 The generator output during the fault. This depends on the faultlocation and type.
3 The fault-clearing time.
4 The post fault transmission system reactance.
5 The generator inertia. The higher the inertia, the slower the rate ofchange in angle. This reduces A1.
6 The generator internal voltage magnitude E . This depends on thefield excitation.
Siva (IIT P) EE549 41 / 61
Numerical Integration Methods
Letdx
dt= f(x, t)
where x is the state vector and f(x, t) is a vector of non linear functions.1 Explicit Methods
1 Euler Method2 Modified Euler Method3 Runge-Kutta Methods
2 Implicit Methods1 Trapezoidal Rule
Siva (IIT P) EE549 42 / 61
Euler Method
Consider the first-order differential equation.
dx
dt= f (x , t)
with x = x0 at t = t0.
t
x
x0
t0
x1
t1
∆x ∆t
Siva (IIT P) EE549 43 / 61
At x = x0, t = t0, the curve can be approximated by its tangent having aslope
dx
dt
∣∣∣∣x=x0
= f (x0, t0)
Therefore,
∆x =dx
dt
∣∣∣∣x=x0
∆t
The value of x at t = t1 = t0 + ∆t is given by
x1 = x0 + ∆x = x0 +dx
dt
∣∣∣∣x=x0
∆t
This has to be repeated till the time reaches the final simulation time.
Since it considers only the firs derivative of x , it is referred to as afirst order method.
∆t has to be small to achieve accuracy.
Since it uses only the first order information, it may introduce error.
Siva (IIT P) EE549 44 / 61
Modified Euler Method
The standard Euler method results in inaccuracies because it uses thederivative at the beginning of the interval.
The modified Euler method tries to overcome this issue by using theaverage of the derivatives at the two ends.
It consists of the following steps.
1 Predictor step
xp1 = x0 +dx
dt
∣∣∣∣x=x0
∆t
2 Corrector step
xc1 = x0 +1
2
(dx
dt
∣∣∣∣x=x0
+dx
dt
∣∣∣∣x=xp1
)∆t
This process has to be repeated until the desired accuracy or the finalsimulation time.
Siva (IIT P) EE549 45 / 61
Runge-Kutta (R-K) Methods
Euler and the modified Euler method require smaller time steps.
R-K methods approximate the Taylor series solution. However they donot need derivatives higher than the first.
R-K methods use the effectiveness of higher derivatives by severalevaluations of the first derivative.
They are classified based on the number of evaluations.
Siva (IIT P) EE549 46 / 61
Second order R- K Method
The value of x at t = t0 + ∆t is
x1 = x0 + ∆x = x0 +k1 + k2
2
wherek1 = f (x0, t0)∆t
k2 = f (x0 + k1, t0 + ∆t)∆t
In general,
xn+1 = xn +k1 + k2
2
wherek1 = f (xn, tn)∆t
k2 = f (xn + k1, tn + ∆t)∆t
Siva (IIT P) EE549 47 / 61
Fourth order R- K Method
The value x at n + 1st step is
xn+1 = xn +1
6(k1 + 2k2 + 2k3 + k4)
wherek1 = f (xn, tn)∆t
k2 = f (xn +k12, tn +
∆t
2)∆t
k3 = f (xn +k22, tn +
∆t
2)∆t
k4 = f (xn + k3, tn + ∆t)∆t
Siva (IIT P) EE549 48 / 61
k1 = (slope at the beginning of time step)∆t
k2 = (first approximation to slope at mid step)∆t
k3 = (second approximation to slope at mid step)∆t
k4 = (slope at the end of step)∆t
∆x =1
6(k1 + 2k2 + 2k3 + k4)
This is equivalent to considering upto fourth derivative terms in the Taylorseries expansion.
Siva (IIT P) EE549 49 / 61
Stability of Explicit Methods
Explicit methods calculate x at any time step from the knowledge of thevalues of x at previous time steps.
They are not numerically stable.
They require smaller time steps.
For the stiff systems (Stiffness is the ratio of the largest to smallesttime constants), they blow up unless a small time step is used.
Stiffness can also be found by the ratio of the largest to smallesteigenvalues of the linearized system.
Siva (IIT P) EE549 50 / 61
Implicit Methods
Consider the differential equation
dx
dt= f (x , t)
with x = x0 and t = t0.
x1 = x0 +
∫ t1
t0
f (x , τ)dτ
Implicit methods approximate the integral.
The simplest implicit integration method is the trapezoidal rule.
The trapezoidal rule approximates the integral by trapezoids.
Siva (IIT P) EE549 51 / 61
Trapezoidal Rule
t
f (x , t)
f (x0, t0)
f (x1, t1)
t1∆tt0
Siva (IIT P) EE549 52 / 61
The trapezoidal rule is
x1 = x0 +∆t
2[f (x0, t0) + f (x1, t1)]
In general
xn+1 = xn +∆t
2[f (xn, tn) + f (xn+1, tn+1)]
xn+1 appears on both sides.
Implicit methods find x using x at the previous time step as well ascurrent value.
Since the current value is unknown, an implicit equation must besolved.
Siva (IIT P) EE549 53 / 61
Stability of Implicit Methods
Implicit methods are numerically stable
For the stiff systems, implicit methods suffer from accuracy but notnumerical stability.
Implicit methods work well with larger time steps.
For systems where time steps are limited by numerical stability ratherthan accuracy, implicit methods are better.
Siva (IIT P) EE549 54 / 61
Example
A 50 Hz synchronous generator having inertia constant H = 5 sec and adirect axis reactance X ′d = 0.3 p.u. is connected to an infinite bus througha transformer and a double circuit line. The network is purely reactive. Thesynchronous generator is delivering real power P = 0.8 p.u. and reactivepower Q = 0.074 p.u. to the infinite bus of 1.0 p.u at steady state.
0.2
E ′d
X ′d = 0.3
V∞ = 1∠0◦
X = 0.3
X = 0.3F
A solid three phase fault occurs at point F and the fault is cleared by openingthe faulted line.
1 Determine the critical clearing angle and the critical fault clearingtime using numerical integration.
2 Check the above angle using the Equal Area Criterion.
Siva (IIT P) EE549 55 / 61
The current flowing into the infinite bus is
I =S∗
V ∗=
0.8− 0.074
1∠0◦= 0.8− 0.074
The reactance between E ′d and V∞ before the fault is
X1 = 0.3 + 0.2 +0.3 + 0.3
2= 0.65
The direct axis transient internal voltage is
E ′d = V∞ + X1I = 1∠0◦ + 0.65× (0.8− 0.074) = 1.17∠26.38◦
Pmax1 =E ′dV∞X1
=1.17× 1
0.65= 1.8
δ0 = 26.38◦
Siva (IIT P) EE549 56 / 61
The reactance during the fault is
X2 =∞
Pmax2 = 0
The reactance after the fault is
X3 = 0.3 + 0.2 + 0.3 = 0.8
Pmax3 =1.17× 1
0.8= 1.46
Siva (IIT P) EE549 57 / 61
The swing equation can be written as follows:
dω
dt=ωs
2H(Pm − Pe)
dδ
dt= (ω − ωs)
The general formula for the second order R-K method
ωn+1 = ωn +
(k1ω + k2ω
2
)δn+1 = δn +
(k1δ + k2δ
2
)where
k1ω =ωs
2H(Pm − Pmax sin(δn))∆t
k1δ = (ωn − ωs)∆t
k2ω =ωs
2H(Pm − Pmax sin(δn + k1ω))∆t
k2δ = (ωn + k1δ − ωs)∆t
Siva (IIT P) EE549 58 / 61
Figure: Swing Curve
Siva (IIT P) EE549 59 / 61
The fault was applied at t = 1 sec.
∆t = 0.05 sec.
The simulation was carried till tf = 5 sec.
It is found that the critical clearing time tcr = 0.2 sec.
δcr =ωs
4HPmt
2cr + δ0
δcr ≈ 55◦
By using the equal area criterion
cos δcr =Pm(δmax − δ0) + Pmax3 cos δmax
Pmax3
Pmax3 = 1.4625
δmax = π − sin−1(Pm
Pmax3) = 2.3887
δcr ≈ 54◦
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Drawbacks of the classical model
1 The dynamics of rotor field winding and the damper winding on thegenerator are totally neglected in the classical model. However, theycan effect the stability of a system significantly.
2 In the classical model, the internal voltage behind the transient reac-tance was assumed to be constant. This is not true since the rotorfiled current is controlled through an exciter and automatic voltageregulator (AVR). Their dynamics have to be included.
3 In the classical model, Pm is assumed to be constant. But Pm dependson speed governor and turbine dynamics. Their dynamics need to beconsidered.
4 Dynamic loads like induction motors, synchronous motors, power elec-tronic devices do affect the stability. In the classical model, they werenot taken into consideration.
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